The formula for calculation of n-factor $(n_{f})$ for a disproportionation reaction is given as:

n_{f} = (n_{f})_{oxidation} + (n_{f})_{reduction}

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\frac{1}{n_{f}} = \frac{1}{(n_{f})_{oxidation}} + \frac{1}{(n_{f})_{reduction}}

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n_{f} = (n_{f})_{oxidation}^{2} + (n_{f})_{reduction}^{2}

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n_{f} = (n_{f})_{oxidation} - (n_{f})_{reduction}

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Solution

The correct option is B $\frac{1}{n_{f}} = \frac{1}{(n_{f})_{oxidation}} + \frac{1}{(n_{f})_{reduction}}$
The formula for calculation of n-factor $(n_{f})$ for disproportionation reaction:
$n_{f} = \frac{(n_{f})_{oxidation} \times (n_{f})_{reduction}}{(n_{f})_{oxidation} + (n_{f})_{reduction}}$

Taking reciprocal both sides

$\frac{1}{n_{f}} = \frac{(n_{f})_{oxidation} + (n_{f})_{reduction}}{(n_{f})_{oxidation} \times (n_{f})_{reduction}}$
$\frac{1}{n_{f}} = \frac{(n_{f})_{oxidation}}{(n_{f})_{oxidation} \times (n_{f})_{reduction}} + \frac{(n_{f})_{reduction}}{(n_{f})_{oxidation} \times (n_{f})_{reduction}}$
Hence,

$\frac{1}{n_{f}} = \frac{1}{(n_{f})_{oxidation}} + \frac{1}{(n_{f})_{reduction}}$

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